Bergman Space Question

**MattMan** · September 5th 2010, 01:20 PM

Let $G=\{z \in \field{C} :0<|z|<1\}$ show that every $f \in L_a^2(G)$ has a removable singularity at $z=0$ .

I'm quite certain this is an easy question, but I'm lacking the background for the appropriate proof. Any help?

**MattMan** · September 7th 2010, 05:25 PM

Since no one apparently knows this, let me go ahead and type this:

z=0 is a removable singularity of $f \in L_a^2(G)$ if there exists a holomorphic function g definied on $G \cup \{0\}$ such that $f(z)=g(z), \forall z \in G$ . Hence let $g=f, \forall f \in G$ . f is holomorphic on $G \cup \{0\}$ , since f is holomorphic on G.

Am I missing something? It seems that by definition of being holomorphic on G, you're holomorphic at {0}, no?

**Jose27** · September 11th 2010, 09:21 PM

Originally Posted by MattMan

Let $G=\{z \in \field{C} :0<|z|<1\}$ show that every $f \in L_a^2(G)$ has a removable singularity at $z=0$ .

I'm quite certain this is an easy question, but I'm lacking the background for the appropriate proof. Any help?

This is a nice problem, unfortunately I don't have a full solution yet.

It's easy to see that f can't have a pole (because then $f(z)=\frac{g(z)}{z^m}$ with $g(0)\neq 0$ and $\frac{1}{z^m} \in L^2(G)$ iff $m\leq 0$ ), but I haven't been able to deduce the case where f has an essential singularity (It's informally clear since f takes on (almost) the whole plane an infinite number of times in any neighbourhood of 0, but the proof eludes me)

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